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Charles Proteus Steinmetz.

Engineering mathematics; a series of lectures delivered at Union college online

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hence, the percentage loss is,



and this is an extreme for the value of current i, given by
hence,



= 0;

iy~v~

or,

Pi ri 2 - =Q and i == \l~ = 5 amperes, ... (8)

and the output is P = ei = l 1,500 watts. The loss is, P = P i +
ri 2 - = 2P t - = 1000 watts; that is, the i 2 r loss or variable loss, is
equal to the constant loss P t -. The percentage loss is,

A = p7= =0.087 = 8.7 per cent,

and the maximum efficiency thus is,

1- A -0.913 = 91 .3 per cent.

102. Usually, when the problem is given, to determine
those values of x for which T/ is an extreme, '*/ cannot be expressed
directly as function of x, y=f(x), as was done in examples
(4) and (5), but y is expressed as function of some other quan-
ties, y=f(u, v . .), and then equations between u, v . . and x
are found from the conditions of the problem, by which expres-
sions of x are substituted for u, v . ., as shown in the following
example :

Example 6. There is a constant current to through a cir-
cuit containing a resistor of resistance r . This resistor r



156 ENGINEERING MATHEMATICS.

is shunted by a resistor of resistance r. What must be the
resistance of this shunting resistor r, to make the power con-
sumed in r, a maximum? (Fig. 56.)

Let i be the current in the shunting resistor r. The power
consumed in r then is,

P = ri 2 (9)

The current in the resistor r is ioij and therefore the
voltage consumed by r is ro(iQ i), and the voltage consumed
by r is ri, and as these two voltages must be equal, since both



w




FIG. 56. Shunted Resistor.



resistors are in shunt with each other, thus receive the same
voltage,



and, herefrom, it follows that,



Substituting this in equation (9) gives,



and this power is an extreme for = ; hence :

(r + r
hence,



(12)



that is, the power consumed in r is a maximum, if the resistor
r of the shunt equals the resistance r .



MAXIMA AND MINIMA. 157

The current in r then is, by equation (10),

and the power is,



4 '

103. If, after the function y=f(x) (the equation (11) in
example (6) ) has been derived, the differentiation ~7~ = is

immediately carried out, the calculation is very frequently
much more complicated than necessary. It is, therefore,
advisable not to differentiate immediately, but first to simplify
the function y=f(x).

If y is an extreme, any expression differing thereform by
a constant term, or a constant factor, etc., also is an extreme.
So also is the reciprocal of y, or its square, or square root, etc.

Thus, before differentiation, constant terms and constant
factors can be dropped, fractions inverted, the expression
raised to any power or any root thereof taken, etc.

For instance, in the preceding example, in equation (11),



(r+r ) 2 '

the value of r is to be found, which makes P a maximum.
If P is an extreme,

r



which differs irom P by the omission of the constant factor
r 2 io 2 , also is an extreme.
The reverse of yi,

2/2=

is also an extreme. (3/2 is a minimum, where y\ is a maximum,
and inversely.)

Therefore, the equation (11) can be simplified to the form :

2 r 2



158 ENGINEERING MATHEMATICS.

and, leaving out the constant term 2r , gives the final form,

2/3 = r+y (13)

This differentiated gives,

dr ~ r 2 ~ ' '
hence,



104. Example 7. From a source of constant alternating
e.m.f. e, power is transmitted over a line of resistance r and
reactance XQ into a non-inductive load. What must be the
resistance r of this load to give maximum power?

If i current transmitted over the line, the power delivered
at the load of resistance r is

P = ri 2 (14)



The total resistance of the circuit is r+ro; the reactance
is XQ' } hence the current is

* - = (15)






and, by substituting in equation (14), the power is

re 2



if P is an extreme, by omitting e 2 and inverting,



is also an extreme, and likewise,

r
2/2 = r+

is an extreme.



MAXIMA AND MINIMA. 159

Differentiating, gives:



dr r 2

(17)



Wherefrom follows, by substituting in equation (16),
P=



(18)



Very often the function y=f(x) can by such algebraic
operations, which do not change an extreme, be simplified to
such an extent that differentiation becomes entirely unnecessary,
but the extreme is immediately seen; the following example
will serve to illustrate :

Example 8. In the same transmission circuit as in example
(7), for what value of r is the current i a maximum?

The current i is given, by equation (15),



Dropping e and reversing, gives,
Squaring, gives,



dropping the constant term Xo 2 gives

2/ 3 = (r+r ) 2 ; (19)

taking the square root gives



160 ENGINEERING MATHEMATICS.

dropping the constant term r gives

y 5 = r] (20)

that is, the current i is an extreme, when y 5 = r is an extreme,
and this is the case for r-=0 and r = oo : r = gives,



(21)



as the maximum value of the current, and r = oo gives



as the minimum value of the current.

With some practice, from the original equation (1), imme-
diately, or in very few steps, the simplified final equation can
be derived.

105. In the calculation of maxima and minima of engineer-
ing quantities x, y, by differentiation of the function y=f(x),
it must be kept in mind that this method gives the values of
x, for which the quantity y of the mathematical equation y =f(x)
becomes an extreme, but whether this extreme has a physical
meaning in engineering or not requires further investigation;
that is, the range of numerical values of x and y is unlimited
in the mathematical equation, but may be limited in its engineer-
ing application. For instance, if x is a resistance, and the
differentiation of y=f(x) leads to negative values of x, these
have no engineering meaning; or, if the differentiation leads
to values of x, which, substituted in y=f(x), gives imaginary, or
negative values of y, the result also may have no engineering
application. In still other cases, the mathematical result
may give values, which are so far beyond the range of indus-
trially practicable numerical values as to be inapplicable.
For instance :

Example 9. In example (8), to determine the resistance
r, which gives maximum current transmitted over a trans-
mission line, the equation (15),



MAXIMA AND MINIMA. 161

immediately differentiated, gives as condition of the extremes:

di^ __ 2(r+r )

dr 2{(r + r ) 2 + z 2 1 V(r + r ) 2 +atf

hence, either r + r = 0; ....... (22)

or, (r+r ) 2 +x 2 = oo . . ..... (23)



the latter equation gives r = oo; hence i = 0, the minimum value
of current.

The former equation gives

r - r , ....... . (24)

as tne value of the resistance, which gives maximum current,
and the current would then be, by substituting (24) into (15),



(25)



The solution (24), however, has no engineering meaning,
as the resistance r cannot be negative.

Hence, mathemetically, there exists no maximum value
of i in the range of r which can occur in engineering, that is,
within the range, 0< r< oo.

In such a case, where the extreme falls outside of the range
of numerical values, to which the engineering quantity is
limited, it follows that within the engineering range the quan-
tity continuously increases toward one limit and continuously
decreases toward the other limit, and that therefore the two
limits of the engineering range of the quantity give extremes.
Thus r = gives the maximum, r = oo the minimum of current.

1 06. Example 10. An alternating-current generator, of
generated e.m.f. e = 2500 volts, internal resistance r = 0.25
ohms, and synchronous reactance o = 10 ohms, is loaded by
a circuit comprising a resistor of constant resistance r = 20
ohms, and a reactor of reactance x in series with the resistor
r. What value of reactance x gives maximum output?

If i = current of the alternator, its power output is

P = n 2 = 20i 2 ; . (26)



162 ENGINEERING MATHEMATICS.

the total resistance is r + TO = 20.25 ohms; the total reactance
is x+xo^lQ+x ohms, and therefore the current is



(27)



and the power output, by substituting (27) in (26), is

P. = 2Q X 25QQ2 (98)

(r + r 2 ) + (z+zo) 2 20.25 2 + (10+z) 2 '

Simplified, this gives

2 + (x+x ) 2 ; . . . (29)



hence,
and



(30)



that is, a negative, or condensive reactance of 10 ohms. The
power output would then be, by substituting (30) into (28),

re 2 20+2500 2

watts = 305 kw. . . (31)



20.25 2

If, however, a condensive reactance is excluded, that is,
it is assumed that x >0, no mathematical extreme exists in the
range of the variable x, which is permissible, and the extreme
is at the end of the range, x = 0, and gives

P=TZ . "!* 9 = 245kw. (32)



107. Example n. In a 500-kw. alternator, at voltage
6 = 2500, the friction and windage loss is P M , = 6 kw., the iron
loss P; = 24 kw., the field excitation loss is P f =6 kw., and the
armature resistance r=0.1 ohm. At what load is the efficiency
a maximum?



MAXIMA AND MINIMA. 163

The sum of the losses is:

P = P w; + P t .+P / + n; 2 -36,000+0.1i 2 . . . . (33)

The output is



(34)
hence, the efficiency is

P U ei 2500,:



or, simplified,

hence,

and,




_ nn
= 600 amperes, . (36)



and the output, at which the maximum efficiency occurs, by
substituting (36) into (34), is



that is, at three times full load.

Therefore, this value is of no engineering importance, but
means that at full load and at all practical overloads the
maximum efficiency is not yet reached, but the efficiency is
still rising.

1 08. Frequently in engineering calculations extremes of
engineering quantities are to be determined, which are func-
tions or two or more independent variables. For instance,
the maximum power is required which can be delivered over a
transmission line into a circuit, in which the resistance as well
as the reactance can be varied independently. In other
words, if

y-Mt>) ...... (37)



164 ENGINEERING MATHEMATICS.

is a function of two independent variables u and v, such a
pair of values of u and of v is to be found, which makes y a
maximum, or minimum.

Choosing any value UQ, of the independent variable u,
then a value of v can be found, which gives the maximum (or
minimum) value of y, which can be reached for U = UQ. This
is done by differentiating y=f(uo,v), over v, thus:



(38)



From this equation (38), a value,

v=fi(u ), ....... (39)

is derived, which gives the maximum value of y, for the given
value of UQ, and by substituting (39) into (38),

....... (40)



is obtained as the equation, which relates the different extremes
of y, that correspond to the different values of UQ, with UQ.
Herefrom, then, that value of u is found which gives the
maximum of the maxima, by differentiation :

0. (41)



Geometrically, y=f(u,v) may be represented by a surface
in space, with the coordinates y, u, v. y =f(uo,v), then, represents
the curve of intersection of this surface with the plane UQ =
constant, and the differentation gives the maximum point
of this intersection curve. 2/=/2(wo) then gives the curve
in space, which connects all the maxima of the various inter-
sections with the UQ planes, and the second differentiation
gives the maximum of this maximum curve y=f2(uo), or the
maximum of the maxima (or more correctly, the extreme of
the extremes).

Inversely, it is possible first to differentiate over u, thus,

fe*U() ....... (42)

du



MAXIMA AND MINIMA. 165

and thereby get



....... (43)

as the value of u, which makes y a maximum for the given
value of V = VQ, and substituting (43) into (42),

....... (44)



is obtained as the equation of the maxima, which differentiated
over VQ, thus,

dftfyo)

j =0, (45)

gives the maximum of the maxima.

Geometrically, this represents the consideration of the
intersection curves of the surface with the planes v = constant.

The working of this will be plain from the following example :

109. Example 12. The alternating voltage e = 30,000 is
impressed upon a transmission line of resistance ro = 20 ohms
and reactance .TO = 50 ohms.

What should be the resistance r and the reactance x of the
receiving circuit to deliver maximum power?

Let i = current delivered into the receiving circuit. The
total resistance is (r+r ); the total reactance is (X+XQ); hence,
the current is



(46)



The power output is

P = n2; ....... (47)

hence, substituting (46), gives

re 2



(a) For any given value of r, the reactance x, which gives

., dP

maximum power, is derived by -r- = 0.



P simplified, gives 2/i = (x + .r ) 2 ; hence,

=0 and X=-X Q ; . . . (49)



166 ENGINEERING MATHEMATICS.

that is, for any chosen resistance r, the power is a maximum,
if the reactance of the receiving circuit is chosen equal to that
of the line, but of opposite sign, that is, as condensive reactance.
Substituting (49) into (48) gives the maximum power
available for a chosen value of r, as :



TP 2

....... (50)



(r+roF
or, simplified,

_(r+r ) 2 = r +^-

2/2 r 2/3 r ,

hence,

^3_ rp 2

and by substituting (51) into (50), the maximum power is,



(b) For any given value of x, the resistance r, which gives

dP n
maximum power, is given by -7- = 0.

P simplified gives,

yi=



_ ,
dr



(53)



which is the value of r, that for any given value of x, gives
maximum power, and this maximum power by substituting
(53) into (48) is,



[r + Vr 2 + (a; + z ) 2 ] 2 + (^ + ^o) 2

c 2

; (54)



MAXIMA AND MINIMA. 167

which is the maximum power that can be transmitted into a
receiving circuit of reactance .r.

The value of x, which makes this maximum power PQ the

highest maximum, is given by ^~ = 0-
P simplified gives



and this value is a maximum for (x+xo)=0; that is, for

x=-x (55)

NOTE. If x cannot be negative, that is, if only inductive
reactance is considered, x = Q gives the maximum power, and
the latter then is

e 2



the same value as found in problem (7), equation (18).

Substituting (55) and (54) gives again equation (52), thus,

P -*-
max 4r '

no. Here again, it requires consideration, whether the
solution is practicable within the limitation of engineering
constants.

With the numerical constants chosen, it would be



Q

i = ~ := 750 amperes,



168 ENGINEERING MATHEMATICS.

and the voltage at the receiving end of the line would be
e t = i^fi +X 2 = 750V20 2 + 50 2 = 40,400 volts;

that is, the voltage at the receiving end would be far higher
than at the generator end, the current excessive, and the efficiency
of transmission only 50 per cent. This extreme case thus is
hardly practicable, and the conclusion would be that by the
use of negative reactance in the receiving circuit, an amount
of power could be delivered, at a sacrifice of efficiency, far
greater than economical transmission would permit.

In the case, where capacity was excluded from the receiv-
ing circuit, the maximum power was given by equation (56) as



-=6100 k\V.




in. Extremes of engineering quantities x, y, are usually
determined by differentiating the function,



and from the equation,



- ......... <



deriving the values or x, which make y an extreme.

Occasionally, however, the equation (58) cannot be solved
for x, but is either of higher order in x, or a transcendental
equation. In this case, equation (58) may be solved by approx-
imation, or preferably, the function,

, = ^ . (59)

dx,

is plotted as a curve, the values of x taken, at which 2 = 0,

that is, at which the curve intersects the X-axis. For instance :

Example 13. The e.m.f. wave of a three-phase alternator,

as determined by oscillograph, is represented by the equation,

e = 36000|sin 0-0.12 sin (30-2.3) -23 sin (50-1.5) +

0.13 sin (70-6.2)}. . . . (60)



MAXIMA AND MINIMA. 169

This alternator, connected to a long-distance transmission line,
gives the charging current to the line of

i = 13.12cos (0-0.3) -5.04 cos (3d- 3.3) -18.76 cos (50-3.6)

+ 19.59 cos (70- 9.9) .... (61)

(see Chapter III, paragraph 95).

What are the extreme values of this current, and at what
phase angles 6 do they occur?

The phase angle 0, at which the current i reaches an extreme
value, is given by the equation




FIG. 57.
Substituting (61) into (62) gives,

di

z = ^p -13.12 sin (0-0.3) +15. 12 sin (30 -3.3) +93.8 sin

(50- 3.6) -137.1 sin (70- 9.9) = 0. . . . (63)

This equation cannot be solved for 0. Therefore z is
plotted as function of by the curve, Fig. 57, and from this
curve the values of taken at which the curve intersects the
zero line. They are:

= 1; 20; 47 78; 104; 135; 162.



170 ENGINEERING MATHEMATICS.

For these angles 0, the corresponding values of i are calculated
by equation (61), and are:

t'o=+9; -1; +39; -30; +30; -42; +4amperes.

The current thus has during each period 14 maxima, of
which the highest is 42 amperes.

112. In those cases, where the mathematical expression
of the function y=f(x) is not known, and the extreme values
therefore have to be determined graphically, frequently a greater
accuracy can be reached by plotting as a curve the differential
of y=f(x) and picking out the zero values instead of plotting
yf(x), and picking out the highest and the lowest points.
If the mathematical expression of y=f(x) is not known, obvi-
ously the equation of the differential curve z= (64) is usually

(tx

not known either. Approximately, however, it can fre-
quently be plotted from the numerical values of yf(x) t as
follows :

If xi t x 2 , x 3 . . . are successive numerical values of x,
and 2/i; 2/2, 2/3 ^ nc corresponding numerical values of y,

approximate points of the differential curve z = -r are given
by the corresponding values:

as abscissas :



as



,. 2/2-2/1 2/3-2/2 2/4-2/3

ordmates : : ; . . .



113. Example 14, In the problem (1), the maximum permea-
bility point of a sample of iron, of which the (B, 3C curve is given
as Fig. 51, was determined by taking from Fig. 51 corresponding

/D

values of (B and 3C, and plotting [JL=, against (B in Fig. 52.

l)C

A considerable inaccuracy exists in this method, in locating
the value of (B, at which /i is a maximum, due to the flatness
of the curve, Fig. 52.



MAXIMA AND MINIMA.



171



The successive pairs of corresponding values of (B and 3C,
as taken from Fig. 51 are given in columns 1 and 2 of Table I.

TABLE I.



<B

Kilo Lines.


5C


<B

V=K


J/z


i
(B








370






1


1.76


570


+ 200


0.5


2


2.74


730


160


1.5


3


3.47


865


135


2.5


4


4.06


985


120


3.5


5


4.59


1090


105


4.5


6


5.10


1175


85


5.5


7


5.63


1245


70


6.5


8


6.17


1295


50


7.5


9


6.77


1330


35


8.5


10


7.47


1340


+ 10


9.5


11


8.33


1320


-20


10.5


12


9.60


1250


70


11.5


13


11.60


1120


130


12.5


14


15.10


930


190


13.5


15


20.7


725


205


14.5



In the third column of Table I is given the permeability,

/n

//= , and in the fourth column the increase of permeabilitv,

3C

per (B=l, ^//; the last column then gives the value of (B, to
which J/i corresponds.

In Fig. 58, values J/t are plotted as ordinates, with (B as
abscissas. This curve passes through zero at (B = 9.95.

The maximum permeability thus occurs at the approximate
magnetic density (B = 9.95 kilolines per sq.cm., and not at (B =
10.2, as was given by the less accurate graphical determination
of Fig. 52, and the maximum permeability is / = 1340.

As seen, the sharpness of the intersection of the differential
curve with the zero line permits a far greater accuracy than
feasible by the method used in instance (1).

114. As illustration of the method of determining extremes,
some further examples are given below:



172



ENGINEERING MATHEMATICS.



Example 15. A storage battery of ft =80 cells is to be
connected so as to give maximum power in a constant resist-
ance r = 0.1 ohm. Each battery cell has the e.m.f. e = 2.1
volts and the internal resistance r = 0.02 ohm. How must
the cells be connected?

Assuming the cells are connected with x in parallel, hence

ft

- in series. The internal resistance of the battery then is



n






= JT ohms, and the total resistance of the circuit is -^r + r.



A,.



Kilq-line



\




13 13 14 1



-100-



FIG. 58. First Differential Quotient of (B,// Curve

7? 7?

The e.m.f. acting on the circuit is e^ since - cells of e.m.f.

Ju JC



are in series. Therefore, the current delivered by the battery



is,



and the power which this current produces in the resistance
r, is,

P=ri 2 =



rn 2 e 2



MAXIMA AND MINIMA. 173

This is an extreme, if



is an extreme, hence,

Tx
and

x



In^o

is, x=+
\ r



/^o

=Vv= 4 '



that is, x=\l~ == 4 cells are connected in multiple, and

= A / = 20 cells in series.
x \r

115. Example 16, In an alternating-current transformer the
loss of power is limited to 900 watts by the permissible temper-
ature rise. The internal resistance of the transformer winding
(primary, plus secondary reduced to the primary) is 2 ohms,
and the core loss at 2000 volts impressed, is 400 watts, and
varies with the 1.6th power of the magnetic density and there-
fore of the voltage. At what impressed voltage is the output
of the transformer a maximum?

If e is the impressed e.m.f. and i is the current input, the
power input into the transformer (approximately, at non-
inductive load) is P = ei.

If the output is a maximum, at constant loss, the input P
also is a maximum. The loss of power in the winding is
ri* = 2i 2 .

The loss of power in the iron at 2000 volts impressed is
400 watts, and at impressed voltage e it therefore is

/ e V 6

(2000) X400 >

and the total loss in the transformer, therefore, is



174 ENGINEERING MATHEMATICS.

herefrom, it follows that,



and, substituting, into P=ei-:



Simplified, this gives,



"2000^'
and, differentiating,

dy 3.6e 2 ' 6

de ""2000 1 - 6 '

and

(-^Y M -125

\2000/

Hence,

1.15 and e = 2300 volts,



which, substituted, gives



P = 2300V450-200X1.25 = 32.52 kw. .

116. Example 17. In a 5-kw. alternating-current transformer,
at 1000 volts impressed, the core loss is 60 watts, the i 2 r loss
150 watts. How must the impressed voltage be changed,
to give maximum efficiency, (a) At full load of 5-kw; (6) at
half load?

The core loss may be assumed as varying with the 1.6th
power of the impressed voltage. If e is the impressed voltage,

i = - is the current at full load, and ii = is the dux-rent at
e e

half load, then at 1000 volts impressed, the full-load current is

5000

= 5 amperes, and since the i 2 r loss is 150 watts, this gives



MAXIMA AND MINIMA. 175

the internal resistance of the transformer as r = 6 ohms, and
herefrom the i 2 r loss at impressed voltage e is respectively,

150 X10 6 37.5 X10 6

n 2 = - and nf = 5 - watts.

Since the core loss is 60 watts at 1000 volts, at the voltage e
it is



watts.

The total loss, at full load, thus is
P P

FL=

and at half load it is



L, = Pi + rii 2 = 60 X



Simplified, this gives



\100Q

hence, differentiated,



' 6 +0.625 Xl0 6 Xe- 2 ;




1000 1 ' 6

e 3 - 6 = 3.125 X10 6 X1000 1 ' 6 = 3.125 XlO 108 ;

e 3 - 6 = 0.78125 X 10 6 X 1000 1 - 6 = 0.78125 X 10 10 - 8 ;
hence, e = 1373 volts for maximum efficiency at full load,
and e = 938 volts for maximum efficiency at half load.

117. Example 18. (a) Constant voltage e = 1000 is im-
pressed upon a condenser of capacity 0=10 mf., through a
reactor of inductance L = 100 mh., and a resistor of resist-
ance r = 40 ohms. What is th* maximum value of the charg-
ing; current?




176 ENGINEERING MATHEMATICS.

(b) An additional resistor of resistance r' = 210 ohms is
then inserted in series, making the total resistance of the con-
denser charging circuit, r = 250 ohms. What is the maximum
value of the charging current?

The equation of the charging current of a condenser, through
a circuit of low resistance, is (" Transient Electric Phenomena
and Oscillations/ 7 p. 61) :



where



and the equation of the charging current of a condenser, through
a circuit of high resistance, is (" Transient Electric Phenomena
and Oscillations," p. 51),



where



Substituting the numerical values gives:

(a) i = 10.2 - 200 < sin 980 t]

(b) i = 6.667 { -*- -***>'}.

Simplified and differentiated, this gives:

( a ) 2=^ = 4.9 cos 980^-sin 980^=0;

hence tan 980^ = 4.9

980^ = 68.5 =1.20

1,20 -tn*

t==



MAXIMA AND MINIMA. 177

hence,



= 0.00092 sec.,

and, by substituting these values of t into the equations of the
current, gives the maximum values:

1.20 + nx

(a) i = lQe - 4^~ = 7.83 - 64n = 7.83X0.53" amperes;


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